# CBSE Class 10 Math Question Paper 2016 | Q9

###### CBSE Maths Board Paper Solution | 2-Mark Questions | Arithmetic Progression Class 10 Maths Questions

The given question is a medium difficulty 2-mark question from the chapter Arithmetic Progressions. Concept tested: Applying the formula to compute the nth term of an arithmetic progression. The question appeared in Section B of the 2016 CBSE class 10 board paper.

Question 9: The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term

## NCERT Solution to Class 10 Maths

### Explanatory Answer | CBSE 2016 Maths Paper Q9

The nth term of AP, an = a1 + (n - 1) d
where a1 is the first term and d is the common difference

Given Data: a4 = 0
i.e., a4 = a1 + (4 - 1) d = 0
or a1 + 3d = 0
or a1 = -3d ..... (1)

##### To Prove: a25 = 3(a11)

a25 = a1 + (25 - 1 ) d = a1 + 24d ..... (2)
Substitute a1 = -3d in eqn (2)
a25 = -3d + 24d = 21d ..... (3)

a11 = a1 + (11 - 1) d = a1 + 10d ..... (4)
Substitute a1 = -3d in eqn (4)
a11 = -3d + 10d = 7d ..... (5)

Compare equations (3) and (5)
Equation (3): a25 = 21d
Equation (5): a11 = 7d
21d = 3 × 7d
Or a25 = 3 × a11

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